Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M/F$ be a quadratic extension of number fields, with Galois group $G=\{1,\tau\}$. Consider the following unitary group $$U_1(R)=\{z\in (R\otimes_FM)^\times :zz^\tau=1\},$$ where $R$ is an $F$-algebra. If we have a character $$\eta:U_1(F)\backslash U_1(\mathbb{A}_F)\to\mathbb{C}^\times,$$ then we can get a Hecke character $\chi$ on $\mathbb{A}_M$ by setting $\chi(z):=\eta\left(\frac{z}{z^\tau}\right)$.

Question: What would be a necessary and sufficient condition for a Hecke character $\chi$ on $\mathbb{A}_M$ to descend to a character on $U_1(\mathbb{A}_F)$? At first, I thought that anti-Galois invariance (i.e. $(\chi^\tau)^{-1}=\chi)$ would be enough, but I don't see how to prove this.

share|cite|improve this question
By Galois invariance, you mean "anti-Galois-invariance", I guess -- i.e. that $\chi(x x^\tau) = 1$? – David Loeffler May 30 '12 at 19:39
@DavidLoeffler Yes, thank you for pointing this out. – M Turgeon May 30 '12 at 20:23
Hilbert's theorem 90 seems to be the mechanism to prove your surmise. – paul garrett Jul 23 '12 at 18:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.